Question 11.7: The 45-kVA, 220-V, 60-Hz, six-pole, three-phase synchronous ......

a. (i) The motor will operate at its synchronous speed which can be found from Eq. 4.41

n_{\mathrm{s}}=\left(\frac{120}{\text { poles }}\right) f_{\mathrm{e}} \quad r/min \quad \quad \quad (4.41)

n_{\mathrm{s}}=\left(\frac{120}{\text { poles }}\right) f_{\mathrm{e}}=\left(\frac{120}{6}\right) 60=1200  \mathrm{r} / \mathrm{min}

(ii) As seen in Chapter 5, the field current can be determined from the generated voltage. For motor operation,

\hat{E}_{\text {af }}=\hat{V}_{\mathrm{a}}-j X_{\mathrm{s}} \hat{I}_{\mathrm{a}}=1.0-j 0.836 \times 1.0=1.30 \angle-39.9^{\circ}

where V_a has been chosen as the reference phasor. Thus the field current is

I_{\mathrm{f}}=1.30 \times 2.84=3.69 \mathrm{~A}

Note that we have chosen to solve for E_{af} in per unit. A solution in real units would have of course produced the same result.

b. (i) When the frequency is reduced from 60 Hz to 50 Hz, the motor speed will drop from 1200 r/min to 1000 r/min.

(ii) Let us again consider the equation for the generated voltage

\hat{E}_{\mathrm{af}}=\hat{V}_{\mathrm{a}}-j X_{\mathrm{s}} \hat{I}_{\mathrm{a}}

where here we will assume that the equation is written in real units. As the inverter frequency is reduced from 60 Hz, the inverter voltage will drop proportionally since the inverter maintains constant V/Hz. Thus we can write

V_{\mathrm{a}}=\left(\frac{\omega_{\mathrm{m}}}{\omega_{\mathrm{m} 0}}\right) V_{\mathrm{a}0}

where the subscript 0 is used to indicate a 60-Hz value as found in part (a). Reactance is also proportional to frequency and thus

X_{\mathrm{s}}=\left(\frac{\omega_{\mathrm{m}}}{\omega_{\mathrm{m} 0}}\right) X_{\mathrm{s} 0}

The generated voltage is proportional to both the motor speed (and hence the frequency) and the field current, and thus we can write

E_{\mathrm{af}}=\left(\frac{\omega_{\mathrm{m}}}{\omega_{\mathrm{m} 0}}\right)\left(\frac{I_{\mathrm{f}}}{I_{\mathrm{f} 0}}\right) E_{\mathrm{af}0}

Finally, if we recognize that, to operate at rated torque and unity power factor under this reduced frequency condition, the motor armature current will have to be equal to the value found in part (a), i.e., I_{\mathrm{a}}=I_{\mathrm{a} 0}, we can write the generated voltage equation as

\left(\frac{\omega_{\mathrm{m}}}{\omega_{\mathrm{m} 0}}\right)\left(\frac{I_{\mathrm{f}}}{I_{\mathrm{f} 0}}\right) \hat{E}_{\mathrm{af} 0}=\left(\frac{\omega_{\mathrm{m}}}{\omega_{\mathrm{m} 0}}\right) \hat{V}_{\mathrm{a} 0}-j\left(\frac{\omega_{\mathrm{m}}}{\omega_{\mathrm{m} 0}}\right) X_{\mathrm{s} 0} \hat{I}_{\mathrm{a} 0}

or

\left(\frac{I_{\mathrm{f}}}{I_{\mathrm{f} 0}}\right) \hat{E}_{\mathrm{af0}}=\hat{V}_{\mathrm{a} 0}-j X_{\mathrm{s} 0} \hat{I}_{\mathrm{a}0}

Since the subscripted quantities correspond to the solution of part (a), they must satisfy

\hat{E}_{\mathrm{af0}}=\hat{V}_{\mathrm{a}0}-j X_{\mathrm{s}0} \hat{I}_{\mathrm{a}0}

and thus we see that we must have I_{\mathrm{f}}=I_{\mathrm{f} 0}. In other words, the field current for this operating condition is equal to that found in part (a), or I_{\mathrm{f}}=3.69 \mathrm{~A}.